Nut insert for repair

[Chuck in E. TN]

 John(Doubleboost),
Very neat repair you demonstrated in Sunday Nite Nightcap # 66 (

0iPSQ&index=1). Is there a formula or rule of thumb that lead you to choose 7mm to bore for a M4 nut? Any correlation that can be used for such repairs using SAE nuts?
All the best to you and Debs.
Chuck

[studders]

I've used that method a few times in wood, never occurred to me to use it in aluminumium. One to remember.

[studders]

To do this we need to calculate the diameter of a nut ACROSS THE POINTS .........

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Doubtless that was the very method John used to arrive at 7mm.

I favour the 'guess it and whack it' approach.

[awemawson]

I claim prior invention on this as per my posting in my Fanuc Wire EDM rebuild thread in October 2014 


(Actually the idea was suggested to me by a poster on the MIG welding forum)

[mklotz]

I favour the 'guess it and whack it' approach.

I'm sure you do.

[bertie_bassett]

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Could you not just measure the nut??

[sparky961]

I'm sure I heard John go through that formula on the video.  I just couldn't understand him.

[krv3000]

I think he added ten to it to get the factor of 7

[doubleboost]

Sorry Lads
No magic formula I measured across the flats and just went with that
The flat bottomed hole it the important bit

I will have a play with different size nuts & post a bit video

Any further DRO installs will be made from steel
 John

[vtsteam]

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Could you not just measure the nut??

 


And then of course apply the technical term "slightly smaller than" to arrive at an exact value.

[RobWilson]

 

 


And then of course apply the technical term "slightly smaller than" to arrive at an exact value.

   PRICELESS !   


Rob 

[RobWilson]

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Just checked above with  my book " Math to build on a book for those who build "  close enough  ,but  the book did not cover the "slight smaller factor" 


Rob